DOTTORATO DI RICERCA IN ECONOMIA POLITICA (XII CICLO) Econometrics test (28/11/2011)
Nome:
1. Say if the following statements are unambiguously true (TRUE), unambiguously false (FALSE) or impossible to classify the way they are stated (CAN’T SAY). Write the moti- vations to your answers only in the space provided. Answers with no motivations will not be considered.
(a) If x is a n × 1 column vector, then the square symmetric matrix xx
0has full rank.
TRUE FALSE CAN’T SAY
(b) If ε
t∼ W N (0, Σ), the covariance matrix Σ is diagonal.
TRUE FALSE CAN’T SAY
(c) The White test for heteroskedasticity can be seen as a Lagrange Multiplier (LM) type test.
TRUE FALSE CAN’T SAY
(d) The parameters for the logit model cannot be estimated by the OLS method.
TRUE FALSE CAN’T SAY
(e) The GLS estimator is generally more efficient than the OLS estimator.
TRUE FALSE CAN’T SAY
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2. All the production functions in a sample of N = 500 enterprises are given by the follow- ing Cobb-Douglas equation
Y
i= AL
αiK
iβ,
in which we assume A = 1. The variable Y
iis the total amount of production by the i-th firm and the production factors are labour (L
i> 0) and capital (K
i> 0). The total sample is split into two subsamples containing N
1= 50 large firms and N
2= 450 small- and medium-size firms. Other data are provided in the following Table.
subsample n
n
X
i=1
y2i
n
X
i=1
l2i
n
X
i=1
κ2i
n
X
i=1
liyi n
X
i=1
κiyi n
X
i=1
liκi
large 50 200000 80000 100000 64000 80000 40000 SMEs 450 300900 50000 80000 33000 42000 20000
yi= ln Yi, li= ln Liand κi= ln Kifor each i.
The OLS estimates for the entire sample are ˆ α = 0.512 and ˆ β = 0.508, while the OLS estimates for the second subsample are ˆ α
1= 0.5, and ˆ β
1= 0.4. The SSR for the whole sample is 28035.
(a) Compute the OLS estimator of α and β for the subsample 1.
(b) Calculate a suitable test for the presence of a break between different categories of enterprises. The SSR in sample 1 is 8000, while the SSR in sample 2 is 17000.
Test: Distribution: Test stat.:
Result: ACCEPT REJECT
(c) Test the hypothesis of constant return to scale in subsample 2.
H
0:
Test: Distribution: Test stat.:
Result: ACCEPT REJECT
3. Let x
t, y
tand w
tbe the 1-year, 5-year, and 10-year US Treasury Constant Maturity rates, respectively. Their time path is depicted in Figure 1 and a few results are reported in Tables 1 and 2. A dummy variable crisis is set in 2008:12 to account for some relevant effects of the subprime crisis.
(a) Write Model 2 in ECM form (d
tcontains the constant, the trend and the dummy).
∆w = d
t+
(b) Test, if possible, the restrictions imposed by Model 1 to Model 2.
NO: it is not possible to carry out a test statistic (provide a motivation)
YES: it is possible (carry out the test)
Test: Distribution: Test stat.:
Result: ACCEPT REJECT
(c) Provide some comments on the estimates.
2
0 1 2 3 4 5 6 7
2000 2002 2004 2006 2008 2010
x y w
Figure 1: US rates
Table 1: Model 1
OLS, using observations 1999:01-2011:03 (T = 147) Dependent variable: w
coefficient std. error t-ratio p-value --- const 1.44620 0.13638 10.60 1.03e-19 ***
time -0.00049 0.00033 -1.470 0.1439
crisis -0.27283 0.10264 -2.658 0.0088 ***
x -0.24464 0.01494 -16.37 1.63e-34 ***
y 0.99242 0.02640 37.59 1.11e-75 ***
Mean dependent var 4.432381 S.D. dependent var 0.906030 Sum squared resid 1.426852 S.E. of regression 0.100241 R-squared 0.988095 Adjusted R-squared 0.987759
F(4, 142) 2946.362 P-value(F) 1.7e-135
Log-likelihood 132.0857 Akaike criterion -254.1714 Schwarz criterion -239.2193 Hannan-Quinn -248.0962
rho 0.807056 Durbin-Watson 0.381588
Breusch-Godfrey test for autocorrelation up to order 4
Test statistic: LMF = 73.364391, p-value = P(F(4,138) > 73.3644) = 3.32e-33
Alternative statistic: TRˆ2 = 99.982630, p-value = P(Chi-square(4) > 99.9826) = 9.92e-21
Ljung-Box of order 4
Test statistic Q’ = 188.83, p-value = P(Chi-square(4) > 188.83) = 9.46e-40
Test for ARCH of order 4
Test statistic: LM = 42.1923, p-value = P(Chi-square(4) > 42.1923) = 1.52183e-08
Test for normality of residuals:
x Jarque-Bera test = 0.347175, p-value 0.840644
Augmented Dickey-Fuller test for residuals including 12 lags - sample size 134
Test with constant: tau_c(1) = -3.19336, asymptotic p-value 0.02041
Test with constant and trend: tau_ct(1) = -4.28773, asymptotic p-value 0.003225
KPSS test for residuals (without trend) T = 147 - Lag truncation parameter = 5
Test statistic = 0.144621 (critical values: 0.349 [10%], 0.464 [5%], 0.737[1%])
3
Table 2: Model 2
OLS, using observations 1999:01-2011:03 (T = 147) Dependent variable: x
coefficient std. error t-ratio p-value ---
const 0.22712 0.09685 2.345 0.0204 **
time 0.00001 0.00018 0.065 0.9479
crisis -0.46914 0.05599 -8.379 5.26e-14 ***
w_1 0.82859 0.04622 17.93 5.75e-38 ***
x -0.28031 0.03259 -8.602 1.49e-14 ***
x_1 0.24014 0.03013 7.970 5.17e-13 ***
y 1.00289 0.02740 36.61 3.00e-73 ***
y_1 -0.83420 0.05143 -16.22 7.22e-34 ***
Mean dependent var 4.432381 S.D. dependent var 0.906030 Sum squared resid 0.390249 S.E. of regression 0.052986 R-squared 0.996744 Adjusted R-squared 0.996580
F(7, 139) 6078.499 P-value(F) 1.7e-169
Log-likelihood 227.3741 Akaike criterion -438.7481 Schwarz criterion -414.8247 Hannan-Quinn -429.0278
rho 0.076166 Durbin’s h 1.109458
Godfrey test for autocorrelation up to order 4
Test statistic: LMF = 0.935694, p-value = P(F(4,135) > 0.935694) = 0.445
Alternative statistic: TRˆ2 = 3.965524, p-value = P(Chi-square(4) > 3.96552) = 0.411
Ljung-Box of order 4
Test statistic Q’ = 3.41761, p-value = P(Chi-square(4) > 3.41761) = 0.491
Test for ARCH of order 4
Test statistic: LM = 6.71653, p-value = P(Chi-square(4) > 6.71653) = 0.151648
Test for normality of residuals:
Jarque-Bera test = 3.38441, p-value 0.184113
Dickey-Fuller test for residuals sample size 146
Test with constant: tau_c(1) = -11.2196, p-value 2.334e-17
Test with constant and trend: tau_ct(1) = -11.1819, p-value 2.702e-16
KPSS test for e2 (without trend) T = 147 - Lag truncation parameter = 5
Test statistic = 0.0890061 (critical values: 0.349[10%], 0.464[5%], 0.737[1%])
